A boat is moving away from an observation tower. When the boat is 50 m from the foot of the tower, the angle of depression of the boat from the top is 60°. After 8 seconds, the angle of depression becomes 30°. Assuming still water, what is the approximate speed of the boat in km/hr?

Introduction / Context:
This problem combines height and distance with speed, time and distance concepts. A boat moves away from a fixed observation tower. As it moves, the angle of depression from the top of the tower to the boat changes from 60 degrees to 30 degrees over a known time interval. The aim is to use trigonometry to determine the horizontal distances corresponding to these angles and then compute the boat’s speed from the change in distance over time.

Given Data / Assumptions:

• At the first instant, the boat is 50 m from the foot of the tower horizontally.
• Angle of depression at that instant = 60 degrees.
• After 8 seconds, the angle of depression becomes 30 degrees.
• The boat is moving directly away from the tower in still water in a straight line.
• The tower is vertical and the ground or water surface is horizontal.
• Angles of depression from the top equal angles of elevation from the boat to the top.

Concept / Approach:
First, we use the initial angle of 60 degrees and distance 50 m to find the height of the tower using tan 60 = height / base. Then, using the same tower height and the second angle of 30 degrees, we find the new horizontal distance from the tower. Once we have both distances, we calculate how far the boat has travelled in 8 seconds. Finally, we convert this speed from metres per second to kilometres per hour using the conversion factor 1 m/s = 3.6 km/hr.

Step-by-Step Solution:
Let the height of the tower be h. Initially, base distance x1 = 50 m and angle of elevation is 60 degrees, so tan 60 = h / 50. Thus sqrt(3) = h / 50, giving h = 50√3 m. After 8 seconds, let the new horizontal distance be x2 and angle of elevation be 30 degrees. Then tan 30 = h / x2, so 1 / sqrt(3) = 50√3 / x2. Therefore x2 = 150 m. Distance travelled by the boat in 8 seconds = x2 − x1 = 150 − 50 = 100 m. Speed in m/s = distance / time = 100 / 8 = 12.5 m/s. Convert to km/hr: 12.5 × 3.6 = 45 km/hr.

Verification / Alternative check:
We can check the tower height with the second distance: using x2 = 150, tan 30 = h / 150 = 1 / sqrt(3), so h = 150 / sqrt(3) = 50√3, consistent with the first calculation. Also, a boat covering 100 m in 8 seconds is traveling at 12.5 m/s, which is a realistic speed for a fast boat, and 12.5 × 3.6 = 45 km/hr aligns exactly with the chosen option.

Why Other Options Are Wrong:

• 33 km/hr and 42 km/hr: These are lower than the correct conversion from 12.5 m/s and do not match the exact calculations.
• 50 km/hr and 60 km/hr: These are higher than the derived speed and would require the boat to cover more than 100 m in 8 seconds, which contradicts the geometry and time data.
• Any value not equal to 45 km/hr fails to satisfy both angle conditions and the 8 second interval simultaneously.

Common Pitfalls:
One common mistake is to misinterpret the initial 50 m as the slant distance instead of the horizontal distance. Another is forgetting that angle of depression equals angle of elevation for the corresponding line of sight. Students may also incorrectly convert from metres per second to kilometres per hour. Always carefully distinguish between vertical height, horizontal distance and the line of sight forming the hypotenuse.

Final Answer:
The approximate speed of the boat is 45 km/hr.